Categories for the Working Mathematician says
Definition. A monad $T = \langle T, \eta, \mu\rangle$ in a category $X$ consists of a functor $T \colon X \to X$ and two natural transformations $$ \eta \colon I_X \Rightarrow T, \quad \mu \colon T^2 \Rightarrow T $$ which make the following diagrams commute: $$ \require{AMScd} \begin{CD} T^3 @>{T \mu}>> T^2 \\ @V{\mu T}VV @VV{\mu}V \\ T^2 @>>{\mu}> T \end{CD} \qquad \begin{CD} IT @>{\eta T}>> T^2 @<{T \eta}<< TI \\ @| @VV{\mu}V @| \\ T @= T @= T \end{CD} $$ (Original images of the diagrams)
Given $\eta \colon I_X \Rightarrow T$, i.e., $\eta$ is a natural transformation from identity functor $I_X$ to functor $T$, I sometimes have the illusion that $\eta$ should just be $T$, because $\eta(I_X) = T$ and $T \circ I_X = T$. So is $\eta$ the same as $T \circ (-)$?
I wonder how to explain whether that is correct or incorrect?
Thanks.
The equation $\eta(I_X) = T$ doesn't really make sense. A natural transformation isn't a machine that takes in functors and produces new functors, its a machine that takes in objects $A \in X$ and produces morphisms $\eta_A: A \to T(A)$.
For a general functor $T$, $A$ and $T(A)$ may have no relationship whatsoever, so in that sense the morphisms that make up $\eta$ are a non-trivial piece of data - $\eta$ is not "just" equal to any one thing in particular.
A simple example to play with is a category $X$ coming from a monoid $M$, considered as a category with one object $*$ such that $Hom(*,*) = M$. Then functors $T$ are simply monoid endomorphisms $T \in End(M)$. A natural transformation $\eta: I_X \to T$ is then just a singleton collection $\eta = \{\eta_*\}$ of one morphism for the object $*$, which is by definition some element $n \in M$. The naturality condition says that for all $x \in M$, one has $T(x)\cdot n = n\cdot x$.
In the special case that $T = I$ is the identity functor, the equation becomes $xn = nx$ so that choices of $\eta$ correspond to central elements $n \in Z(M)$. To extend this example to an actual monad you need to assume $n$ has a left inverse which then plays the role of $\mu$ - this is because the righmost diagram requires $x = (\mu_* \circ \eta_*)(x)$ for all $x \in M$.)
NOTE: Sorry I realize now that $\eta_*, \mu_*$ looks kind of like "pushforward" notation, but here I mean the natural transformation evaluated at the object $*$ of a one-object category.